Answer
The numbers are $x=-3$ ; $y=0$ ; $z=2$
Work Step by Step
Let us consider the given system of equations and mark the equations as shown below:
$\frac{x+3}{2}-\frac{y-1}{2}+\frac{z+2}{4}=\frac{3}{2}$ (A)
$\frac{x-5}{2}+\frac{y+1}{3}-\frac{z}{4}=-\frac{25}{6}$ (B)
$\frac{x-3}{4}-\frac{y+1}{2}+\frac{z-3}{2}=-\frac{5}{2}$ (C)
Simplify the first equation by taking the L.C.M and the equation becomes:
$\begin{align}
& 2\left( x+3 \right)-2\left( y-1 \right)+z+2=6 \\
& 2x+6-2y+2+z+2=6 \\
& 2x-2y+z+10=6
\end{align}$
Now, subtract 10 from both sides:
$\begin{align}
& 2x-2y+z+10-10=6-10 \\
& 2x-2y+z=-4
\end{align}$
Simplify the second equation by taking the L.C.M and the equation becomes:
$\begin{align}
& 6\left( x-5 \right)+4\left( y+1 \right)-3z=-50 \\
& 6x-30+4y+4-3z=-50 \\
& 6x+4y-3z-26=-50
\end{align}$
By adding 26 to both sides we get:
$\begin{align}
& 6x+4y-3z-26+26=-50+26 \\
& 6x+4y-3z=-24
\end{align}$
Simplify the third equation by taking the L.C.M and the equation becomes:
$\begin{align}
& x-3-2\left( y+1 \right)+2\left( z-3 \right)=-10 \\
& x-3-2y-2+2z-6=-10 \\
& x-2y+2z-11=-10
\end{align}$
Add 11 to both sides to get:
$\begin{align}
& x-2y+2z-11+11=-10+11 \\
& x-2y+2z=1
\end{align}$
The system of equations now becomes:
$2x-2y+z=-4$
$6x+4y-3z=-24$
$x-2y+2z=1$
Mark equations as shown below:
$2x-2y+z=-4$ (I)
$6x+4y-3z=-24$ (II)
$x-2y+2z=1$ (III)
By multiplying equation (III) by −2 and eliminate x equations (I) and (II): we get,
$\begin{align}
& \text{ }2x-2y+\text{ }z=-4 \\
& \underline{-2x+4y-4z=-2} \\
& \text{ 2}y-3z\text{ }=-6 \\
\end{align}$ (IV)
By multiplying equation (I) by −6 and eliminating x from equations (I) and (II): we get
$\begin{align}
& \text{ }6x+\text{ }4y-\text{ }3z=\text{ }-24 \\
& \underline{-6x+12y-12z=\text{ }-\text{ }6} \\
& \text{ }16y\text{ }-\text{ 15}z=-\text{ }30 \\
\end{align}$ (V)
Multiply equation (IV) by −8 and then add equations (IV) and (VI) to get the value of z:
$\begin{align}
& \text{ }16y-15z=-30 \\
& \underline{-16y+24y=\text{ }48} \\
& \text{ }9z=18 \\
\end{align}$ (VI)
By dividing equation (V) by 9 we get:
$\begin{align}
& \frac{9z}{9}=\frac{18}{9} \\
& \text{ }z=2
\end{align}$
Substitute the value of z in equation (IV):
$\begin{align}
& 2y-3\left( 2 \right)=-6 \\
& 2y-6=-6
\end{align}$
Now, adding 6 to both sides, we get
$\begin{align}
& 2y-6+6=-6+6 \\
& 2y=0
\end{align}$
It implies $y=0.$
Now, putting the values of y and z in equation (III) to get the value of x:
$\begin{align}
& x-2\left( 0 \right)+2\left( 2 \right)=1 \\
& x+4=1
\end{align}$
Now, subtract 4 from both sides of the equation:
$\begin{align}
& x+4-4=1-4 \\
& x=-3
\end{align}$
Hence, the values are $x=-3$, $y=0,$ and $z=2$.
